python - 使用 Pandas 或 Numpy 的 n 维滑动窗口

Question

如何使用 Numpy 或 Pandas 执行与 rollapply(...., by.column=FALSE) 等效的 R(xts)？当给定一个数据帧时，pandas rolling_apply 似乎只能逐列工作，而不是提供向目标函数提供完整(窗口大小)x(数据帧宽度)矩阵的选项。

import pandas as pd
import numpy as np

xx = pd.DataFrame(np.zeros([10, 10]))
pd.rolling_apply(xx, 5, lambda x: np.shape(x)[0]) 

    0   1   2   3   4   5   6   7   8   9
0 NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN
1 NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN
2 NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN
3 NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN
4   5   5   5   5   5   5   5   5   5   5
5   5   5   5   5   5   5   5   5   5   5
6   5   5   5   5   5   5   5   5   5   5
7   5   5   5   5   5   5   5   5   5   5
8   5   5   5   5   5   5   5   5   5   5
9   5   5   5   5   5   5   5   5   5   5

所以发生的事情是 rolling_apply 依次沿着每一列向下移动，并在每一列下方应用一个 5 长度的滑动窗口，而我想要的是滑动窗口每次都是一个 5x10 数组，在这种情况下，我会得到一个单列向量(不是二维数组)结果。

Answer 1

我确实找不到一种方法来计算 pandas 中的“广泛”滚动应用程序 文档，所以我会使用 numpy 在数组上获取“窗口化” View 并应用 ufunc 给它。这是一个例子:

In [40]: arr = np.arange(50).reshape(10, 5); arr
Out[40]: 
array([[ 0,  1,  2,  3,  4],
       [ 5,  6,  7,  8,  9],
       [10, 11, 12, 13, 14],
       [15, 16, 17, 18, 19],
       [20, 21, 22, 23, 24],
       [25, 26, 27, 28, 29],
       [30, 31, 32, 33, 34],
       [35, 36, 37, 38, 39],
       [40, 41, 42, 43, 44],
       [45, 46, 47, 48, 49]])

In [41]: win_size = 5

In [42]: isize = arr.itemsize; isize
Out[42]: 8

arr.itemsize 是 8 因为默认 dtype 是 np.int64，你需要它用于以下“窗口” View 习惯用法:

In [43]: windowed = np.lib.stride_tricks.as_strided(arr,
                                                    shape=(arr.shape[0] - win_size + 1, win_size, arr.shape[1]),
                                                    strides=(arr.shape[1] * isize, arr.shape[1] * isize, isize)); windowed
Out[43]: 
array([[[ 0,  1,  2,  3,  4],
        [ 5,  6,  7,  8,  9],
        [10, 11, 12, 13, 14],
        [15, 16, 17, 18, 19],
        [20, 21, 22, 23, 24]],

       [[ 5,  6,  7,  8,  9],
        [10, 11, 12, 13, 14],
        [15, 16, 17, 18, 19],
        [20, 21, 22, 23, 24],
        [25, 26, 27, 28, 29]],

       [[10, 11, 12, 13, 14],
        [15, 16, 17, 18, 19],
        [20, 21, 22, 23, 24],
        [25, 26, 27, 28, 29],
        [30, 31, 32, 33, 34]],

       [[15, 16, 17, 18, 19],
        [20, 21, 22, 23, 24],
        [25, 26, 27, 28, 29],
        [30, 31, 32, 33, 34],
        [35, 36, 37, 38, 39]],

       [[20, 21, 22, 23, 24],
        [25, 26, 27, 28, 29],
        [30, 31, 32, 33, 34],
        [35, 36, 37, 38, 39],
        [40, 41, 42, 43, 44]],

       [[25, 26, 27, 28, 29],
        [30, 31, 32, 33, 34],
        [35, 36, 37, 38, 39],
        [40, 41, 42, 43, 44],
        [45, 46, 47, 48, 49]]])

步幅是沿给定轴的两个相邻元素之间的字节数， 因此 strides=(arr.shape[1] * isize, arr.shape[1] * isize, isize) 表示跳过 5 从 windowed[0] 到 windowed[1] 时跳过 5 个元素 从 windowed[0, 0] 到 windowed[0, 1]。现在你可以在 结果数组，例如:

In [44]: windowed.sum(axis=(1,2))
Out[44]: array([300, 425, 550, 675, 800, 925])